integral of (x-3)/(x^2-12x+36)

Lösung

\int \frac{x - 3}{x ^{2} - 12 x + 36} d x

ln ∣ x - 6 ∣ - \frac{3}{x - 6} + C

Schritte zur Lösung

\int \frac{x - 3}{x ^{2} - 12 x + 36} d x

Wende U-Substitution an

= \int \frac{u}{u ^{2} - 6 u + 9} d u

Vervollständige das Quadrat

u^{2} - 6 u + 9 : (u - 3)^{2}

= \int \frac{u}{( u - 3 ) ^{2}} d u

Wende U-Substitution an

= \int \frac{v + 3}{v ^{2}} d v

Multipliziere aus

\frac{v + 3}{v ^{2}} : \frac{1}{v} + \frac{3}{v ^{2}}

= \int \frac{1}{v} + \frac{3}{v ^{2}} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{v} d v + \int \frac{3}{v ^{2}} d v

\int \frac{1}{v} d v = ln ∣ v ∣

\int \frac{3}{v ^{2}} d v = - \frac{3}{v}

= ln ∣ v ∣ - \frac{3}{v}

Ersetze zurück

= ln ∣ x - 3 - 3 ∣ - \frac{3}{x - 3 - 3}

Vereinfache

= ln ∣ x - 6 ∣ - \frac{3}{x - 6}

Füge eine Konstante zur Lösung hinzu

= ln ∣ x - 6 ∣ - \frac{3}{x - 6} + C

n = 1 \sum \infty \frac{n}{n + 3}

t an g e n t f (x) = (2 + 4 x) (5 - 4 x), at x = 1

t an g e n t y = 2 x^{3} - 2 x^{2} - 2 x

\frac{d y}{d x} + \frac{y}{x} = x^{3}

\frac{d}{d x} (\frac{e ^{a x^{2} + b x + c}}{x})